2i(3-2i)+4-i=-3(3-2i)2i

Solution for 2i(3-2i)+4-i=-3(3-2i)2i equation:

2i(3-2i)+4-i=-3(3-2i)2i

We move all terms to the left:

2i(3-2i)+4-i-(-3(3-2i)2i)=0

We add all the numbers together, and all the variables

2i(-2i+3)-i-(-3(-2i+3)2i)+4=0

We add all the numbers together, and all the variables

-1i+2i(-2i+3)-(-3(-2i+3)2i)+4=0

We multiply parentheses

-4i^2-1i+6i-(-3(-2i+3)2i)+4=0


We calculate terms in parentheses: -(-3(-2i+3)2i), so:

-3(-2i+3)2i

We multiply parentheses

12i^2-18i

Back to the equation:

-(12i^2-18i)


We add all the numbers together, and all the variables

-4i^2+5i-(12i^2-18i)+4=0

We get rid of parentheses

-4i^2-12i^2+5i+18i+4=0

We add all the numbers together, and all the variables

-16i^2+23i+4=0

a = -16; b = 23; c = +4;
Δ = b2-4ac
Δ = 232-4·(-16)·4
Δ = 785
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{785}}{2*-16}=\frac{-23-\sqrt{785}}{-32}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{785}}{2*-16}=\frac{-23+\sqrt{785}}{-32}
$